Function composition of symmetries n, 1/n for FLTSubj: Re: Nonassociative and nondistributive operations Date: 19 Oct 2002 (sci.math) From: Nico Benschop Orgn: Digital Research : Finite & Associative  nhp wrote: > > Could somebody give an example of some "reasonable" operation > > a) not obeying the associative law (i.e. (x*y)*z != x*(y*z)) > b) not obeying the distributive law (i.e. x*(y+z) != x*y + x*z) > c) not obeying neither associative nor distributive law > > Of course one can always introduce some quite arbitrary * and + such > that the laws are not obeyed, but I'm looking for some operations > which are actually used in maths or at least are somehow "reasonable". > > As simple as the laws are, they are (or maybe just because of that) at > the heart of the maths. I think knowing some examples about what the > laws exlude would give further insight into what these laws and > calculus is all about. The distributive law holds for one operation (*) over another (+) if (*) is defined as repetition of (+). Think of the normal arithmetic operations (^) over (*), and (*) over (+) for integers. To get a feeling for how restrictive 'associative' property is, notice that it is necessary for the important concept of 'iteration': e.g.: (aa)a = a(aa) is denoted a^3 only in this associative case. Imagine an order n closure of set S under some operation (.) defined by its n x n composition table. There are n^{n^2} ways to fill such table, of which n! row/col permutations yield isomorphism's. Now try to find the % of such square tables (closures) that are in fact associative. Then you'll find for not too large n already (say n=10) an almost negligible fraction is indeed associative. Meaning that the 'associative' assumption for an operation: semigroup S(.) is a VERY strong restriction (something like 'linear' for networks), implying a lot of 'structure'... For nonassociative "de beer is los" (anything goes;)  one example is exponentiation (^) over (+), such as FLT (additive statement with exponential terms):  NB  http://piazza.iae.nl/users/benschop/sgrpflt.htm (on semigroups & function composition in arithmetic) Function composition of symmetries n, 1/n for FLT
(news:sci.math . . 4apr98)See On discovering the triplets (mod p^k) . . . (doubleclick)
This triplet structure is used in a direct proof of FLT (Fermat's Last Theorem), The smallest example is Z (mod 3^2) = 2* = {2 4 8 7 5 1} = {2 4 1 2 4 1} mod 9. This 3loop holds for each n<>1 in G mod p^k, for each prime p, and all k>0, except for the cubic roots of 1 mod p^k (p=1 mod 6) where the looplength is one: a+1 = 1/a (a<>1), and in a special case where 4 divides p1 (so p=1 mod 4). . . . Moreover, the Hensel lift (of extending a mod p^2 solution to mod p^k for any k>2) can be "broken" by this a+1= 1/a solution, because the Exponent p Distributes over a Sum: (a+1)^p = a^p + 1 = a + 1 (mod p^k, EDS property in Core . : . see my paper on " Triplets ..." , ref[1] on my homepage). The clue, to prove no loop longer than 3 exists in ring Z(+, .) are its two symmetries (= automorphism of order 2), seen as functions, namely: and : Inverse function I(n) = 1/n for multiply ( neutral element '1' ). "Simple comme bonjour", would Fermat have said... ... although for pth powers the first triplet is at p=59, and arithmetic doodling mod 59^2 in those days with his PC (Pascal Calculator = "Pascaline") is a bit unlikely.  Yet, we don't need pth powers to see the triplet structure of arithmetic, as shown above! Moreover, for p=7 the cubic roots occur already, which he very well could have found (for inequality FLT case1 they should be the only solutions mod p^2, which they are NOT). Check out http://www.iaehv.nl/users/benschop/ferm.htm And the same 3loop holds for ALL four possible compositions of these three elementary arithmetic functions C(n), I(n) and the successor S(n)=n+1. Notice that I and C commute, so 4 rather than 3! = 6 such "dfs" functions arise (dual folded successor functions). but requires more powerful function composition (semigroups). As it were .: You need a diamond (sgrp) to cut steel (arithm). For instance the above function is SIC(n)= 1/(n+1) = SCI(n), where function composition is normally from left to right  following this nice notation advocated by Clifford/Preston in the reference work: "The Algebraic Theory of Semigroups" (AMS Survey #7, 1961) You see that . : . looplength three = the number of symmetries + 1 = the number of operations + 1.
( . . I think this is not a coincidence, but a necessity . . ) The extension to inequality for integers follows from a variant of the Exponent p Distributes over a Sum (EDS) property of this solution of FLT case1 in residues ("two terms of a triplet are in Core"  see my homepage ref[1] on the general Triplet structure of Arithmetic mod p^k, not only for pth power residues). Constructive comment is welcome.

( Do not associate this title with Science = Sick ;) ( Rather: an exercise in function composition ) Consider in arithmetic ring Z(+, .) the two basic symmetries n and 1/n as functions, in fact automorphisms of order 2, of Z(+) resp. Z(.): And denote the successor function as: S(n) = n+1. Notice that IC = CI (commute), but S does not commute with I or C. (so indeed SCI = SIC ;) Then compose all three functions in all possible (four) ways: n(SCI) = 1/(1+n) n(CSI) = 1/(1n) n(ISC) = (1+ 1/n) Watch notation: function composition is from left to right. Call such composition a "dfs" (dual folded successor) function. . . . . The third iteration of each dfs function is the identity function E . . . . . . ( nE = n, for all n<>0,1,1 ).
Proof: . . Let F_i be the ith iteration of a function F. Then show: (dfs)_3 = E.
. . . n(SCI)_2 = 1/[1 1/(1+n)] = (1+n)/[1+n 1] = (1+n)/n = 1/n 1 . . . n(SCI)_3 = (1+n) 1 = n . . . QED. Similarly, verify the other three dfs functions to have period 3.
An interesting consequence of this "3loop" property of arithmetic is
that for residues mod p^k (prime p>2, k>1):
. . . . a+1 = 1/b > b = 1/(a+1) (provided division by zero is avoided, so e.g: a+1 <>0, etc.)
Basic triplet example is in G(.) mod 9 = 2* = {2, 4, 8, 7, 5, 1} . .
where 8= 1, 7= 2, 5= 4 (mod 9).
. . . . . 1(SCI)_*: 1/(1+1)=4, 1/(4+1)=7, 1/(7+1)=1, with abc=1.4.7=1 mod 9 Note : maximal period=3 is the number of symmetries +1 (coincidence?)
. . . For this function composition result applied to a proof of FLTcase1, 
Re: proof of Goldbach's Conjecture . . . sci.math9sep99
Subject: Re: proof of Goldbach's Conjecture Author: Nico Benschop Date: 9 Sep 99 07:04:03 0400 (EDT) seminole1215@mydeja.com wrote: > > Why you guys are wasting the time on this problem, I would never > believe anyone in this planet could prove the Goldbach's Conjecture > with 6page, not even with 60page. Much harder, guys. Handwaving > stuff won't help and old tools with which people proved the cases of > (1+n) (n>1) couldn't be used for case (1+1). > I truely believe we need brandnew tool to task it. ...[*] > It might take a couple of decades and even centuries. ...[&] > Re[*]: Not quite: rather USE tools that are wellknown, such as the algebra of function composition [ associative, but not commutative: f(g(x)) =/= g(f(x) ] to solve hard problems in arithmetic. (like on powersums: Fermat, Waring; or primesums: Goldbach;) For instance: the two symmetries of arithmetic: complement n under (+) about "0", and inverse 1/n under (.) about "1", have a fascinating 3loop property that only can be seen under function composition. Namely, call them 'C' and 'I' respectively, and let 'S' be the Peano successor function n > n+1. Now consider function SCI (from left to right apply S first, then C and lastly I), then you'll easily verify that this function 3 times applied in iteration, yields the identity function E: n > n. (do not divide by 0, so some restrictions hold: n<>0, and n<>1 ): So: n(SCI)= 1/(n+1) applied twice more to itself: n(SCI SCI)= 1/{1/(n+1) +1} = (n+1)/{1 + n+1} = (n+1)/n n(SCI SCI SCI)= 1/{(n+1)/n +1} = n/{n1 +n} = n. Funny, magic, how come: a basic 3loop linking the fundamental two symmetries of arithmetic and the Peano successor function ...!? E.g: this is the clue to all solutions of x^p + y^p = z^p mod p^k, namely the "Triplet": a+1=1/b, b+1=1/c, c+1=1/a, with abc=1 mod p^k basic 3loop structure of residue arithmetic mod p^k (prime p>2), (for ALL residues coprime to p) & a 'sideline' to integers: FLTcase1, breaking the Hensellift by taking into account the 'carry': the pth power of a k_digit number (base p) has upto kp digits   the carry makes the difference, for FLT ;) Indeed: an OLD and well known tool (function composition) applied in a NEW way!  That's why I find: mathematicians are sitting on a goldmine (semigroups = associative algebra, especially in_the_finite) without really knowing it... (like Shannon in 1938 suggested to apply Boolean Algebra  a commutative & idempotent form of arithmetic  to the specification and design of combinational logic circuits: Boole's work was some 90 years old 1848: his monograph precursor of "The Laws of Thought" 1854). For sequential logic synthesis (FSM: Finite State Machines = computers) no such fundamental & practical tool has as yet been developped, although its basis: semigroup algebra (=function composition) is already existant since 1928 (Shushkewitch). Re[&]: It need not necessarily take that long: IF we, open_minded engineers, scientists, and yes: some mathematicians  (although the latter have the disadvantage of the specialist/expert: single focus;)  USE the tools already developed by previous generations, ...the older & simpler the better... Number theory without fundamental use of function composition algebra [semigroups, like Z(.) analysed additively, including noncmt, finite, with divisors of zero, ..&c] I would dare to call not quite complete, missing an essential concept. ========= It takes Steel to cut Wood, It takes Diamond to cut Steel... ======================== where: Wood = All Practical Purposes (APP in Science & Engineering) Steel = Arithmetic, Calculus, Set_theory (classical methods) Diamond = Associative Function Compostion (Semigroups)  Ciao, Nico Benschop  http://www.iae.nl/users/benschop http://www.iae.nl/users/benschop/campaign.htm http://www.iae.nl/users/benschop/func.htm http://www.iae.nl/users/benschop/fewago.htm http://www.iae.nl/users/benschop/scimat98.htm Subject: Re: logic, combinations, permutations Date: Thu, 3 Aug 2000 12:48:21 GMT From: Nico Benschop Org: Research Newsgrp: sci.math  Niek Sprakel wrote: > > Is there any coherent, consistent and complete theory which relates > permutations, combinations and logic? > I reckon permutations are embeded among multinomialcoefficients and > propositional logic is closely related to binomialcoefficients. > A good context for these algebra's (of permutations, transformations, sets, arithmetic, combinational and sequential logic, FSM: state machines) is given by the common property of the corresponding operations: associative. Hence: semigroups (= associative algebra of functions) is their common context.  Ciao, Nico Benschop  http://piazza.iae.nl/users/benschop/sgrpflt.htm simple sgrps as FSM: http://piazza.iae.nl/users/benschop/cranksm.dvi integer state machines http://piazza.iae.nl/users/benschop/ism.htm 
 N.F.Benschop  July 1998  